In the adjacent figure ABCD is a square and \Delta A P B is an equilateral triangle. Prove that \Delta A P D \cong \Delta B P C .(Hint: In \Delta A P D and \Delta B P C \quad \overline { A D } = \overline { B C } , \overline { A P } = \overline { B P } and \angle P A D = \angle P B C = 90 ^ { \circ } - 60 ^ { \circ } = 30 ^ { \circ } )

Given ABCD is square \therefore AB=BC=CD=AD \angle A=\angle B=\angle C=\angle D=90^0 Given PDB is equilateral \therefore AB=PD=PB \angle (PAB)=\angle (APB)=\angle (PBA)=60^0 \therefore \angle (DAP)=\angle (DAB)-\angle (PAB) =90-60=30^0 \therefore \angle(CBP)=\angle(CBA)-\angle(PBD) =90-60=30^0 In \Delta le DPA and CPB AD=BC [sides of the square] PA=PB [sides of the equilateral] \angle(DAP)=\angle(CBP)=60^0 \therefore \Delta DPA\equiv\Delta CPB [SAS axion] \Rightarrow \Delta DPA\cong \Delta BPC Hence proved.

Prove that any two sides of a \Delta are together greater than twice the median drawn to the third side.

Given:Triangle ABC in which AD is the median.To prove:AB+AC>2ADProof:In ΔADB and ΔEDCAD=DE [By construction]D is the midpoint BC.[DB=DB]ΔADB=ΔEDC [vertically opposite angles]Therefore ΔADB≅ ΔEDC [By SAS congruence criterion.]AB=ED [Corresponding parts of congruent triangles]InΔAEC,AC+ED>AE[sum of any two sides of a triangle is greater than the third side]AC+AB>2AD [AE=AD+DE=AD+AD=2AD and ED=AB]

In the adjacent figure ABCD is a square and \Delta A P B is an equilateral triangle. Prove that \Delta APD \cong \Delta PPC (Hint : In \Delta APD and \Delta B P C \quad \vec { A D } = \overline { B C } , \overline { A P } = \overline { B P } and \angle P A D = \angle P B C = 90 ^ { \circ } - 60 ^ { \circ } = 30 ^ { \circ } ]

Given Δ ABP is an equilateral triangle.Then, AP = BP (Same side of the triangle)and AD = BC (Same side of square)and, ∠ DAP = ∠ DAB - ∠ PAB = 90° - 60° = 30°Similarly,∠ BPC = ∠ ABC - ∠ ABP = 90° - 60° = 30°∴ ∠ DAP = ∠ BPC∴ Δ APD is congruent to Δ BPC ( SAS proved)In Δ APDAP = AD II as AP = AB (equilateral triangle)We know that ∠ DAP = 30°∴ ∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and ∠ APD is on of the base angles.)= 150°/2 = 75° = ∠ APD = 75°Similarly, ∠ BPC = 75°Therefore, ∠ DPC = 360° - (75°+75°+60°)= ∠ DPC = 150°Now in Δ PDC PD = PC as Δ APD is congruent to Δ BPC∴ Δ PDC is an isosceles triangleAnd ∠ PDC = ∠ PCD = (180° - 150°)/2 or ∠ PDC = ∠ PCD = 15°∠ DPC = 150°; ∠ PDC = 15° and ∠ PCD = 15°

In the given figure, AB=DC and BD=CA . Prove that \Delta ABC\cong DCB .

Given AB=CD BD=CA In \triangle ABC & \triangle DCB BC=BC (Common side) \therefore \ By SSS congurency \triangle ABC\cong \triangle DCB

By applying SAS congruence rule, you want to establish that \triangle PQR \equiv \triangle FED . It is given that PQ=FE and RP=DE . What additional information is needed to establish the congruence?

Given PQR \cong FED In SAS congruence rule two sides and one angle are equal to 2 sides & one included angle of another triangle.Hence \angle P = \angle F And \angle Q \cong \angle E SAS means angle between 2 sides.

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Important questions on Criteria For Triangle Congruence

Q: In right triangle ABC , right angled at C , M is the mid-point of hypotenuse AB . C is joined to M and produced to a point D such that DM = CM . Point D is joined to point B (see the given figure). Show that: (i) \triangle{AMC}\cong \triangle{BMD} \left(ii\right) CM = \dfrac{1}{2}AB

Proof (i) In \Delta AMC and \Delta BMD M is the midpoint \Rightarrow AM = BM \Rightarrow \angle CMA = \angle DMB (vertically opposite angles) CM = DM (given)Hence \Delta AMC \cong \Delta BMD by SAS congruence rule(ii) DC = AB Since \Delta DBC \cong \Delta ACB \Rightarrow DM+CM = AB \Rightarrow CM+CM = AB Since CD = CM+DM \Rightarrow 2CM = AB and CM = DM Hence CM = 1/2 AB

Q: ABC and D B C are two isosceles triangles on the same base B C (see Fig.). Show that \angle ABD=\angle ACD.

Given :- \Delta ABC is isosceles AB = AC __(1) \Delta DBC is isosceles DB = DC __(2)To prove :- \angle ABD = \angle ACD Proof :- let us join AD ln \Delta ABD and \Delta ACD , AB = AC [from eq. (1)] BD = CD [ from eq. (2)] AD = AD [common] \Delta ABD \cong \Delta ACD (sss rule) \angle ABD = \angle ACD (CPCT)Hence proved

Q: In right triangle ABC , right angled at C,M is the mid-point of hypotenuse AB . C is joined to M and produced to a point D such that DM = CM . Point D is joined to point B . Show that CM = \dfrac{1}{2}AB

AM=BM=\dfrac{1}{2}AB \Rightarrow \angle =90° DM=CM \triangle DBM and \triangle BCM \Rightarrow BM=BM ( common ) DM=CM ( given ) \angle DMB\cong \triangle BCM \therefore BM=CM=\dfrac{1}{2}AB Hence, solved.

BROWSE BY DIFFICULTY

$Δ A B C = Δ P Q R$

$Δ C B A = Δ P R Q$

$Δ B A C = Δ R P Q$

$Δ P Q R = Δ B C A$

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Important questions on Criteria For Triangle Congruence

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM=CM. Point D is joined to point B (see the given figure). Show that: (i)△AMC≅△BMD (ii)CM=21​AB

In the adjacent figure ABCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPD≅ΔBPC . (Hint: In ΔAPD and ΔBPCAD=BC,AP=BP and ∠PAD=∠PBC=90∘−60∘=30∘)

Prove that any two sides of a Δ are together greater than twice the median drawn to the third side.

ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that ∠ABD=∠ACD.

In the adjacent figure ABCD is a square and ΔAPB is an equilateral triangle. Prove that ΔAPD≅ΔPPC (Hint : In ΔAPD and ΔBPCAD=BC,AP=BP and ∠PAD=∠PBC=90∘−60∘=30∘]

Plot the points A(2,−3),B(−1,2) and (0,−2) on the graph paper. Draw the triangle formed by reflecting these points in the x-aixs. Are the two triangles congruent?

In the given figure, AB=DC and BD=CA. Prove that ΔABC≅DCB.

By applying SAS congruence rule, you want to establish that △PQR≡△FED. It is given that PQ=FE and RP=DE. What additional information is needed to establish the congruence?

In right triangle ABC, right angled at C,M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM=CM. Point D is joined to point B. Show that CM=21​AB

If AB=QR;BC=PR and CA=PQ, then

ΔABC=ΔPQR

ΔCBA=ΔPRQ

ΔBAC=ΔRPQ

ΔPQR=ΔBCA

More important questions

In right triangle $A B C$ , right angled at $C$ , $M$ is the mid-point of hypotenuse $A B$ . $C$ is joined to $M$ and produced to a point $D$ such that $D M = C M$ . Point $D$ is joined to point $B$ (see the given figure). Show that: $(i) △ A M C ≅ △ B M D$ $(i i) C M = \frac{1}{2} A B$

In the adjacent figure $A B C D$ is a square and $Δ A P B$ is an equilateral triangle. Prove that $Δ A P D ≅ Δ B P C$ .
(Hint: In $Δ A P D$ and $Δ B P C \overline{A D} = \overline{B C}, \overline{A P} = \overline{B P}$ and
$∠ P A D = ∠ P B C = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ})$

Prove that any two sides of a $Δ$ are together greater than twice the median drawn to the third side.

$A B C$ and $D B C$ are two isosceles triangles on the same base $B C$ (see Fig.). Show that $∠ A B D = ∠ A C D .$

In the adjacent figure $A B C D$ is a square and $Δ A P B$ is an equilateral triangle. Prove that $Δ A P D ≅ Δ P P C$
(Hint : In $Δ A P D$ and $Δ B P C A D = \overline{B C}, \overline{A P} = \overline{B P}$ and $∠ P A D = ∠ P B C = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}]$

Plot the points $A (2, - 3), B (- 1, 2)$ and $(0, - 2)$ on the graph paper. Draw the triangle formed by reflecting these points in the x-aixs. Are the two triangles congruent?

In the given figure, $A B = D C$ and $B D = C A$ . Prove that $Δ A B C ≅ D C B$ .

By applying $S A S$ congruence rule, you want to establish that $△ P Q R \equiv △ F E D$ . It is given that $P Q = F E$ and $R P = D E$ . What additional information is needed to establish the congruence?

In right triangle $A B C$ , right angled at $C, M$ is the mid-point of hypotenuse $A B$ . $C$ is joined to $M$ and produced to a point $D$ such that $D M = C M$ . Point $D$ is joined to point $B$ . Show that $C M = \frac{1}{2} A B$

If $A B = Q R; B C = P R$ and $C A = P Q$ , then

$Δ A B C = Δ P Q R$

$Δ C B A = Δ P R Q$

$Δ B A C = Δ R P Q$

$Δ P Q R = Δ B C A$