CAT 1995 | Question: 81

Question

Choose the correct option:

AB is perpendicular to BC and BD is perpendicular to AC. CE bisects the angle C; $\angle A=30^{\circ}$ Then, what is $\angle CED$?

• A. $30^{\circ}$
• B. $60^{\circ}$
• C. $45^{\circ}$
• D. $65^{\circ}$

Answer 1

 

$AB\perp BC$ & $BD \perp AC$

∴$\measuredangle ABC = 90^{\circ}  $ & $\measuredangle BDA = \measuredangle BDC = 90^{\circ}$

Now, $\angle BAC = 30^{\circ}$

∴$\measuredangle ACB = 180^{\circ} - (90^{\circ}+30^{\circ})$

$\measuredangle ACB = 180^{\circ} - 120^{\circ}$

$\measuredangle ACB = 60^{\circ} $

Given that $CE$ bisects $\angle C$ 

$∴ \measuredangle ECD =  \dfrac{60^{\circ}}{2}= 30^{\circ}  $

Then, $\measuredangle CED = 180^{\circ} - (90^{\circ}+30^{\circ}) = 60^{\circ}$

Comments

Answer depends on where we consider the point E to be. If the point lies on line BD, then the above answer by Sukanya is correct. If the point E lies on the line AB, then the answer would be different.

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also, then answer could definitely not be 60 or 65 degrees. If we consider the point E to be on line AB, then angle BDE + angle CED = 60 degrees.

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1. How you got $65^{\circ}$?

2. then, how you got $\angle BDE + \angle CED = 60^{\circ}$ ?

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I'm telling that the answer COULD NOT be 65° (or 60°), if the point E lies on the line AB. That is because angle BDE + angle CED = 60°. Proof : consider the point of intersection of lines CE and BD as O. using basic "sum of interior angles of a triangle's law we can find that angle DBA = angle CEB = 60°.  Therefore angle EOB = 60. So angle. Hence angle EOD = 180-60= 120°. Now consider the triangle EOD. you will find that angle E + angle D = 180-angle O = 180-120 = 60°. That's how I got angle BDE + angle CED = 60°