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

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