Question

In given figure, AD and CE are two altitudes of $△ABC$. Prove that
(i) $△AEF\sim △CDF$
(ii) $△ABD\sim △CBE$
(iii) $△AEF\sim △ADB$
(iv) $△FDC\sim △BEC$

Answer 1

(i) In triangles AEF and CDF, we have

$\angle AEF=\angle CDF={90}^0$ [$\because CE\perp AB$ and $AD\perp BC$]

$\angle AFE=\angle CFD$ [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

$△AEF\sim △CDF$ $\left[Henceproved\right]$

(ii) In ${△}^{\prime }sABD$ and $CBE$, we have

$\angle ABD=\angle CBE=\angle B$ [Common angle]

$\angle ADB=\angle CEB={90}^0$ [$\because AD\perp BC$ and $CE\perp AB$]

Thus, by AA-Criterion of similarity, we have

$△ABD\sim △CBE$ $\left[Henceproved\right]$

(iii) In ${△}^{\prime }sAEF$ and $ADB$, we have

$\angle AEF=\angle ADB={90}^0$ [$\because AD\perp BC$ and $CE\perp AB$]

$\angle FAE=\angle DAB$ [Common angle]

Thus, by AA-criterion of similarity, we have

$△AEF\sim △ADB$ $\left[Henceproved\right]$

(iv) In ${△}^{\prime }sFDC$ and $BEC$, we have

$\angle FDC=\angle BEC={90}^0$ [$\because AD\perp BC$ and $CE\perp AB$]

$\angle FCD=\angle ECB$ [Common angle]

Thus, by AA-criterion of similarity, we have

$△FDC\sim △BEC$ $\left[Henceproved\right]$