Question

In the adjacent figure $△ABC,D$ is the midpoint of $BC$. $DE\perp AB,DF\perp AC$ and $DE=DF$. Show that $△BED\cong △CFD$

Answer 1

From $△BED$ and $△CFD$

$BD=CD$ (D is a midpoit of BC)

$DE=DF$ (Given)

$\angle DEB=\angle DFC$ ($DE\perp AB$ and $DF\perp AC$)

Thus,

$△BED\cong △CFD$ (By RHS)