Question

$\Delta$ABC and $\Delta$DBC 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 $\Delta$ ABP$\cong$ $\Delta$ ACP.

Answer 1

To prove : $△ABP\cong △ACP$

Proof: In $△ABD$ and $△ACD$

$AB=AC$ [Given]

$BD=DC$ [Given]

$AD$ is common

$△ABD\cong △ACD$ [SSS postulate]

$\widehat{BAD}=\widehat{CAD}$ [CPCT]

$\Rightarrow \widehat{BAP}=\widehat{CAP}$

In $△ABP$ and $△ACP,$

$AB=AC$ [Given]

$AP$ is common\]

$\widehat{BAP}=\widehat{CAP}$ [proved]

$\therefore △BAP\cong △CAP$ [By SAS postulates]