Question

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
(i) ΔABD is congruent to ΔACD
(ii) ΔABP is congruent to ΔACP ​

Answer 1

Answer:

given that triangle ABC and triangle DBC are two isosceles triangles.

to prove that triangle ABD is congruent to

triangle ACD

proof: In triangle ABD and ACD

AB=AC ( BY isoscles property .)

AD =AD COMMON

angle ADB =ADC

Triangle ABD congruent ACD.

BD=CD CPCT

NOW,

IN triangles APB and ACP

AB =AC

AP=AP(common)

angle APB= APC =90°Each

triangle ABP congruence triangle ACP by S.A.S

BP=CP CPCT

Hence,

proved .

Step-by-step explanation:

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