Will angle chasing help?

Geometry Level 4

Quadrilateral A B C D ABCD has angles B D A = C D B = 5 0 , \angle BDA=\angle CDB=50^\circ, D A C = 2 0 , \angle DAC=20^\circ, and C A B = 8 0 . \angle CAB =80^\circ. Find B C A \angle BCA (in degrees).


The answer is 60.

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5 solutions

Cuong Phi
Dec 3, 2015

We have AB is external bisector of CAD and DB is internal bisector of CDA. Then, we have CB is external bisector of ACD. So ,we have ACB=(180-60):2=60

That is a very nice solution!

Calvin Lin Staff - 5 years, 6 months ago

can you please do a full calculation of what you mean,it makes it easier for me to follow

Asad Jawaid - 5 years, 5 months ago
Reetun Maiti
Dec 3, 2015

just draw the angular bisector of angle ACD. Let it meet BD in Z. then quad BAZC is cyclic. Then after some simple angle chasing the problem is over.

Cool! Same way!!

Rohit Udaiwal - 5 years, 6 months ago
Adarsh Kumar
Dec 3, 2015

Let us call the intersection point of A C AC and B D BD E E and B C E = θ \angle BCE=\theta and E B C = α \angle EBC=\alpha .Since D E DE is the internal bisector of angle D D we have A E E C = A D D C . \dfrac{AE}{EC}=\dfrac{AD}{DC}. Applying sine-rule in A B E , B E C \bigtriangleup ABE,\bigtriangleup BEC 2 A E = B E sin 8 0 E C sin α = B E sin θ 2 A D D C sin α = sin θ sin 8 0 . 2 AE=\dfrac{BE}{\sin 80^{\circ}}\\ \dfrac{EC}{\sin\alpha}=\dfrac{BE}{\sin \theta}\\ \Longrightarrow 2\dfrac{AD}{DC}\sin\alpha=\dfrac{\sin\theta}{\sin80^{\circ}}. Leave that aside for a second and apply sine-rule in A B D , B D C \bigtriangleup ABD,\bigtriangleup BDC ,we get, 2 A D = B D sin 10 0 D C sin α = B D sin ( 60 + θ ) 2 A D D C sin α = sin ( 60 + θ ) sin 10 0 sin ( 60 + θ ) sin 10 0 = sin θ sin 8 0 sin ( 60 + θ ) = sin θ 2AD=\dfrac{BD}{\sin 100^{\circ}}\\ \dfrac{DC}{\sin\alpha}=\dfrac{BD}{\sin(60+\theta)}\\ \Longrightarrow 2\dfrac{AD}{DC}\sin\alpha=\dfrac{\sin(60+\theta)}{\sin 100^{\circ}}\\ \Longrightarrow \dfrac{\sin(60+\theta)}{\sin 100^{\circ}}=\dfrac{\sin\theta}{\sin 80^{\circ}}\\ \Longrightarrow \sin(60+\theta)=\sin\theta Expanding the left hand side and evaluating we obtain tan θ = 3 \tan\theta=\sqrt{3} , hence θ = 6 0 \theta=60^{\circ} .And done!

Moderator note:

Nice solution.

When applying a formula, it is best to leave it in the recognizable form, so that the reader can quickly follow what you are saying. IE Use A E sin 3 0 \frac{AE}{ \sin 30 ^ \circ } instead of 2 A E 2AE , and the reader will immediately say "Ah yes, that is a sin α \frac{a} { \sin \alpha} .

Sorry for the late response,sir and thank you for the advice,i will definitely be careful about this when I write my next solution :)

Adarsh Kumar - 5 years, 6 months ago

With out loss of generality, let AB=1, and angle BCA=X.
A C D = 180 20 50 50 = 60 Applying Sin Law to triangles ABC, ABD, CBD, we get 1 B C = S i n X S i n 80 , B D 1 = S i n 100 S i n 50 , B C B D = S i n 50 S i n ( 60 + X ) , Equating the product of the left sides of the three equations with the right sides, we get, 1 = S i n 100 S i n 80 S i n X S i n ( 60 + X ) , B u t S i n X S i n ( 60 + X ) = 1 S i n 60 C o t X + C o s 60 S i n 60 C o t X + C o s 60 = S i n 100 S i n 80 . C a l c u l a t i n g , X = 60 \angle ACD=180 - 20 - 50 - 50 = 60 ~ \text{Applying Sin Law to triangles ABC, ABD, CBD, we get}\\ \dfrac 1 {BC}=\dfrac {SinX}{Sin80}, ~~~~~~~\dfrac {BD} 1 =\dfrac {Sin100}{Sin50}, ~~~~~~~~\dfrac{BC}{BD}=\dfrac {Sin50}{Sin(60+X)}, \\ \text{Equating the product of the left sides of the three equations with the right sides, we get,}\\ 1=\dfrac{Sin100}{Sin80}*\dfrac {SinX}{Sin(60+X)}, ~ But ~\dfrac {SinX}{Sin(60+X)}=\dfrac 1{Sin60CotX+Cos60} ~\\ \implies Sin60CotX+Cos60=\dfrac{Sin100}{Sin80}. Calculating, X= \Large ~\color{rted}{60}

Most probably angle DBC will be shorter than 30° hence either DBC could be 10 or 20 (assumption) by substituting 10 you get answer 60 while you have 3 tries and you have two cases hence you will be able to get answer (I know its hilarious)

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