In the above figure- AB-AC and D is the mid-point of BCProve that -i- Is triangle ADBcong triangle ADC- -Give reason-ii- Is angle B-angle C- Why-

Question

Toppr Admin · Accepted Answer

Given-xA0- AB-AC and-xA0-D is midpoint-xA0-xA0-So- BD-DC-x2026-x2026-i-xA0-In -x394-ABD-amp-x394-ACD- we have-xA0-xA0-AB-AC-given-xA0-BD-DC-fromeq-i-AD-AD-common-i-x394-ADB-x2245-x394-ADC -BySSScongruencyrule-xA0-ii-x2220-B-x2220-C-CPCT-

In the above figure, $A B = A C$ and $D$ is the mid-point of $B C$
Prove that (i) Is $△ A D B ≅ △ A D C$ ? (Give reason)
(ii) Is $∠ B = ∠ C$ ? Why?

Given: $A B = A C$ and D is midpoint
So, $B D = D C \dots \dots (i)$
In $Δ A B D & Δ A C D$ , we have
$A B = A C (g i v e n)$
$B D = D C (f r o m e q (i))$
$A D = A D (c o m m o n)$
$(i) Δ A D B ≅ Δ A D C$ [ $B y S S S c o n g r u e n c y r u l e]$
$(i i) ∠ B = ∠ C (C P C T)$

In the above figure, AB=AC and D is the mid-point of BCProve that (i) Is △ADB≅△ADC? (Give reason)(ii) Is ∠B=∠C? Why?

Given: AB=AC and D is midpoint So, BD=DC……(i) In ΔABD&ΔACD, we have AB=AC(given) BD=DC(fromeq(i))AD=AD(common)(i)ΔADB≅ΔADC [BySSScongruencyrule] (ii)∠B=∠C(CPCT)

In the above figure, $A B = A C$ and $D$ is the mid-point of $B C$
Prove that (i) Is $△ A D B ≅ △ A D C$ ? (Give reason)
(ii) Is $∠ B = ∠ C$ ? Why?

Given: $A B = A C$ and D is midpoint
So, $B D = D C \dots \dots (i)$
In $Δ A B D & Δ A C D$ , we have
$A B = A C (g i v e n)$
$B D = D C (f r o m e q (i))$
$A D = A D (c o m m o n)$
$(i) Δ A D B ≅ Δ A D C$ [ $B y S S S c o n g r u e n c y r u l e]$
$(i i) ∠ B = ∠ C (C P C T)$