Reflection of C is on AB

Geometry Level 2

A right angle triangle $ABC$ is such that $AC=24$ cm, $CB=10$ cm and $\angle ACB=90^{\circ}$ . A point $D$ on $CB$ is such that when point $C$ is reflected upon line $AD$ , the reflected point will be on line $AB$ . The length of $CD=\frac{a}{b}$ where $a$ and $b$ are coprime integers. Find the value of $a+b$ .

The answer is 29.

3 solutions

Kee Wei Lee
Apr 28, 2014

Firstly by Pythagoras Theorem, $AB=\sqrt{24^2+10^2}=26$

Now let $C'$ be the relfected point of $C$ by line $AD$ .

With point C' If so, it is clear that triangles $ADC$ and $ADC'$ are congruent. So, $CD=C'D$ .

Considering the area of triangles; (let $[ABC]$ denote the area of triangle $ABC$ )

$[ABC]=[ACD]+[ADB]$

$\Rightarrow \frac{1}{2}(24)(10)=\frac{1}{2}(24)(CD)+\frac{1}{2}(26)(C'D)$

$\Rightarrow \frac{1}{2}(24)(10)=\frac{1}{2}(24)(CD)+\frac{1}{2}(26)(CD)$

$\Rightarrow (24)(10)=(24)(CD)+(26)(CD)$

$\Rightarrow 50CD=240$

$\Rightarrow CD=\frac{24}{5}$

Hence $a+b=24+5=29$

Reflection means we have to draw a perpendicular from point 'c' to line AD (mark it as E) and then to extend it till AB (mark it as F), then CE=EF, hou u can draw C'D in your figure,

Prasanna Kulkarni - 7 years, 1 month ago

Ya you can do that and you'd realise that $F=C'$ . I can draw $C'D$ , it is the reflection of $CD$ .

Kee Wei Lee - 7 years, 1 month ago

Yes, thank u

Prasanna Kulkarni - 7 years, 1 month ago

Did the problem in the very same manner!!

Shreyansh Vats - 7 years, 1 month ago

same way , i hate similarities

math man - 6 years, 10 months ago

I addressed the problem in the same way you did in the beginning, but then i found CD in a different way. Instead of considering the areas of the triangles like you did, i just concentrated on the little triangle C'BD. With Pythagoras' Theorem we'll have: C'D²+C'B²=BD² And since: C'D=CD , C'B=AB-AC'=26-24=2 and BD=CB-CD=10-CD Then: C'D²+C'B²=BD² => CD²+2²=(10-CD)² => CD²+4=100-20CD+CD² => CD=96/20=24/5

tesnim blel - 3 years, 2 months ago

Adrian Neacșu
Apr 29, 2014

If $P$ is the symmetric of $C$ from $AD$ then $\Delta ACD =\Delta APD$ .

Then $AD$ is a bisector. Then by using the bisector theorem we have.

$\frac {DB}{CD} = \frac {AB}{AC}$ .

By adding 1 to the proportion we get

$\frac {BC}{CD} = \frac {AC+AB}{AC}$ .

Which will give us the result.

Deepak Jain
May 4, 2014

Easy question Solve the question just like you

Reflection of C is on AB

The answer is 29.

3 solutions

0 pending reports