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

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## 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)  $$\Delta ABD\sim{\ }\text{ }\Delta CBE$$

(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 }} \\ & [ \therefore CE\bot AB\,\,\rm{and}\,\,AD\bot BC;\,\rm{altitudes}] \\ & \angle APE=\angle CPD \quad \left(\text{Vertically}\,\,\text{opposite}\,\,\text{angles} \right) \\ \Rightarrow\qquad &\Delta AEP\sim \Delta CPD \quad \left( \text{AA}\,\,\text{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 A D B &=\angle C E B=90^{\circ} \\ \angle A B D &=\angle C B E \quad (\text {Common angle }) \\ \Rightarrow\qquad \Delta A B D &\sim \Delta C B E \quad (A A \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 A E P=\angle A D B=90^{\circ} \\ & \angle P A E=\angle B A D \quad (\text { Common angle }) \\ \Rightarrow\quad & \Delta A E P \sim \Delta A D B \end{align}

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

\begin{align} \angle P D C &=\angle B E C=90^{\circ} \\ \angle P C D &=\angle B C E \quad (\text { Common angle }) \\ \Rightarrow\quad \Delta P D C&-\Delta B E C \end{align}

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