In the given figure altitudes AD and CE of triangle ABC intersect each

1.	In the given figure altitudes AD and CE of triangle ABC intersect each other at the pointProve that B.AE×CP=CD×AP
Answer» 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, $\bf:\longmapsto\:In \: \<klux>TRIANGLE</klux> \: PEA \: and \: \triangle \: PDC$ $\rm :\longmapsto\:\angle \: PEA \: = \: \angle \: PDC \: \: \: \: \: \{each \: 90 \degree \}$ $\rm :\longmapsto\:\angle \: APE \: = \: \angle \: CPD \: \: \: \: \: \{vert. \: opp. \: angles\}$ $\rm :\implies\:\triangle \: PEA \: \sim \: \triangle \: PDC \: \: \{AA \: Similarity \: Rule \}$ $\rm :\implies\:\dfrac{EA}{DC} = \dfrac{AP}{CP}$ $\bf\implies \:AE \times CP = CD \times AP$ $\large{\boxed{\boxed{\bf{Hence, Proved}}}}$ 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.

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

Answer»

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,

$\bf:\longmapsto\:In \: \<klux>TRIANGLE</klux> \: PEA \: and \: \triangle \: PDC$

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

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

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

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

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

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

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.