# Ex.6.3 Q7 Triangles Solution - NCERT Maths Class 10

## Question

In Figure, altitudes \(AD\) and \(CE\) of \(\Delta ABC\) intersect each other at the point .Show that:

(i) \(\Delta AEP\sim{\ }\Delta CDP\)

(ii) \(\Delta ABD\sim{\ }\Delta CBE\)

(iii) \(\Delta AEP\sim\Delta ADB\)

(iv) \(\Delta \text{ }PDC\sim\Delta BEC \)

**Diagram**

## Text Solution

(i) **Reasoning:**

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This may be referred to as the \(AA\) similarity criterion for two triangles.

**Steps:**

In \(\Delta AEP\) and \(\Delta CDP\)

\[\begin{align}&\angle AEP = \angle CDP = {90^\circ }\\&\left[ \begin{array}{l}\therefore CE \bot AB\,\,{\rm{and}}{\mkern 1mu} {\mkern 1mu} {\rm{AD}} \bot {\rm{BC}};{\mkern 1mu} \,{\rm{altitudes}}\end{array} \right]\\\\&\angle APE = \angle CPD\\&\;\;\left[ \begin{array}{l}{\rm{Vertically}}\,{\rm{opposite}}{\mkern 1mu} \,{\rm{angles}}\end{array} \right]\\\\& \Rightarrow \Delta AEP\sim\Delta CPD\\&\qquad\left[ {{\text{AA Criterion}}} \right]\end{align}\]

**(ii) Reasoning:**

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This may be referred to as the \(AA\) similarity criterion for two triangles.

**Steps:**

In \(\Delta ABD\,{\rm {and}}\,\Delta CBE\)

\[\begin{align}\angle ADB &= \angle CEB = {90^\circ }\\\angle ABD &= \angle CBE\\ &[ \text{Common angle}]\\\\ \Rightarrow \,\Delta ABD &\sim \Delta CBE\\ [AA&\text{ Criterion}]\end{align}\]

**(iii) Reasoning:**

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This may be referred to as the \(AA\) similarity criterion for two triangles.

**Steps:**

In \(\Delta\rm{ A E P}\) and \(\Delta\rm{ A D B}\)

\[\begin{align}\angle AEP &= \angle ADB = {90^\circ }\\\angle PAE &\!=\! \angle BAD \\ &[ \text{Common Angle}]\\\\ \Rightarrow \quad \Delta AEP &\sim \Delta ADB\end{align}\]

**(iv) Reasoning:**

This may be referred to as the \(AA\) similarity criterion for two triangles.

**Steps:**

In \(\Delta P D C \,{\rm{and}}\, \Delta B E C\)

\[\begin{align}\angle PDC &= \angle BEC = {90^\circ }\\\angle PCD &= \angle BCE \\& [\text{Common Angle}]\\\\ \Rightarrow \quad\Delta PDC &- \Delta BEC\end{align}\]