Question

In the given figure altitudes AD and CE of triangle ABC intersect each other at the point
Prove that
B.AE×CP=CD×AP​

Answer 1

Given :-

In ∆ ABC

• AD is altitude intersecting BC at D

• CE is altitude intersecting AB at E.

• Altitude AD and CE intersecting each other at P.

To Prove :-

• AE × CP = CD × AP

Concept Used :-

• Similarity of triangles.

Proof :-

It is given that

$\rm :\longmapsto\:CE \: \perp \: AB$

$\rm :\implies\:\angle \: CEA \: = \: 90 \degree \:$

$\rm :\implies\:\angle \: PEA \: = \: 90 \degree \:$

Also,

$\rm :\longmapsto\:AD \: \perp \: BC$

$\rm :\implies\:\angle \: ADC \: = \: 90 \degree \:$

$\rm :\implies\:\angle \: PDC \: = \: 90 \degree \:$

Now,

$\bf:\longmapsto\:In \: \triangle \: PEA \: and \: \triangle \: PDC$

$\rm :\longmapsto\:\angle \: PEA \: = \: \angle \: PDC \: \: \: \: \: \{each \: 90 \degree \}$

$\rm :\longmapsto\:\angle \: APE \: = \: \angle \: CPD \: \: \: \: \: \{vert. \: opp. \: angles\}$

$\rm :\implies\:\triangle \: PEA \: \sim \: \triangle \: PDC \: \: \{AA \: Similarity \: Rule \}$

$\rm :\implies\:\dfrac{EA}{DC} = \dfrac{AP}{CP}$

$\bf\implies \:AE \times CP = CD \times AP$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. Pythagoras Theorem :-

• This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

• This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

• This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

• If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.