maths

Asked on December 20, 2019 by **Zameer Chris**

Prove that APCQ is a parallelogram

$∠D=∠B$ (Opposite side of parallelogram.)

$∴DA=BC$ (Opposite side of parallelogram.)

$∠x=∠x$ [bisector of opposite angle.]

$∴$ By $ASA$ congruence rule. $[△ADP≅△QBP]$

then by $CPCT,$ $DP=QB⟶(1)$

$∠x=∠x$ [bisector of opposite angle.]

$△ADP≅△QBP$ [by $ASA$]

$AB=DC$ [opposite side of a parallelogram.]

$AQ+QB=DP+PC$ [$QB=DP$ from $1$]

$AQ=PC⟶(2)$

Hence distance between $AP$ and $QC$ are equal at all place hence proved $AP∥QC⟶(3)$

$∴AQCP$ is parallelogram.

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State the congruency of following pairs of triangles.

In $ΔABC$ and $ΔPQR$, $BC=QR,$ $∠A=90_{∘},∠C=∠R=40_{∘}$ and $∠Q=50_{∘}$.

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Hence, $△BEC≅△DCF$

If the above statement is true then mention answer as 1, else mention 0 if false

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Answer: DC produced bisects BC at right angle.

If true then enter 1 else if False enter 0

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$ΔABD$ and $ΔACD,$

AB $=$ AC (Given)

$∠B=∠C$ (because AB $=$ AC)

and $∠ADB=∠ADC$

Therefore, $ΔABD≅ΔACD(AAS)$

So, $∠BAD=∠CAD(CPCT)$

What is the defect in the above arguments?

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