Question

AD and BE are respectively perpendiculars to BC and AC of triangle ABC. Show that
i) CA×CE =CB×CD ii) CDX×CB = CA×DE​

Answer 1

$\bold{\underline{\underline{Answer:}}}$

Given :

• AD is perpendicular to side BC.
• BE is perpendicular to side AC.
• Δ ABC

To show :

• $\bold{CA\times\:CE=\:CB\times\:CD}$
• $\bold{CD\times\:CB=\:CA\times\:DE}$

Solution :

We first need to prove triangles similar before we show the product sides equal to product of other sides.

In Δ ADC and Δ BEC,

$\bold{m\angle{ADC=m{\angle{BEC=90{\degree}}}}}$ $\bold{\sf{\underbrace{Both\:measures\:90\degree}}}$

Now we got one angle similar so let's try getting one more angle similar and thereby prove the two triangles similar by AA test of similarity.

$\bold{m{\angle{ACD={\angle{BCE}}}}}$

$\bold{\sf{\underbrace{Common\:angle}}}$

We know that if we are able to prove any two angles or sides of two triangles similar, then the two triangles will be similar to each other. Here as we have proved two angles similar we can say that,

Δ ADC ~ Δ BEC

Now, we will write the corresponding sides of the triangles.

$\bold{\dfrac{AD}{BE}}$ = $\bold{\dfrac{DC}{EC}}$ = $\bold{\dfrac{AC}{BC}}$

Arrange the order a bit,

$\bold{\dfrac{DA}{EB}}$ = $\bold{\dfrac{CD}{CE}}$ = $\bold{\dfrac{CA}{CB}}$

Now, if we pay a note to the question, we will find that we just need two pairs. So, how to decide which pairs to go with?

Pretty simple, look at the question and choose the required pairs.

For now, I need to go with, $\bold{\dfrac{CA}{CB}}$ and $\bold{\dfrac{CD}{CE}}$

$\bold{\dfrac{CA}{CB}}$ = $\bold{\dfrac{CD}{CE}}$

Cross multiplying,

$\rightarrow$ $\bold{CA\times\:CE\:=\:CB\times\:CD}$

We successfully showed the first one.

Now similarly we will show (prove) the other one too.

For second, we will consider Δ ABC and Δ DEC.

Side AC = Side DC

$\bold{\angle{ACB={\angle{ECD}}}}$

Side BC = Side ED

•°• Δ ABC and Δ DEC are similar by SAS test of similarity.

Now writing the corresponding sides of the triangle.

$\bold{\dfrac{AB}{DE}}$ = $\bold{\dfrac{BC}{EC}}$ = $\bold{\dfrac{AC}{DC}}$

Here too, we just require two pairs of corresponding sides.

$\bold{\dfrac{BC}{EC}}$ = $\bold{\dfrac{AC}{DC}}$

Cross multiplying,

$\bold{CD\times\:CB=CA\times\:CE}$