In a quadrilateral $ABCD$ sides $AB$ and $CD$ are equal, $\angle A=150^\circ,$ $\angle B=44^\circ,$ $\angle C=72^\circ.$ Perpendicular bisector of the segment $AD$ meets the side $BC$ at point $P.$ Find $\angle APD.$ Proposed by F. Bakharev
Problem
Source: Tuymaada 2003, day 1, problem 2.
Tags: trigonometry, geometry, perpendicular bisector, geometry proposed
05.05.2007 19:02
We have that $ \angle{D} = 94$, and denote $ \angle{PDA} = \angle{PAD} = x$, therefore $ \angle{APD} = 180 - 2x$. Therefore we have that: 1) $ \angle{CDP} = 94 - x$, so $ \angle{CPD} = x + 14$, 2) $ \angle{BAP} = 150 - x$, so $ \angle{APB} = x - 14$. Now from the sin law we have: 1) $ \frac {AP}{\sin{44}} = \frac {AB}{\sin{(x - 14)}}$. 2) $ \frac {DP}{\sin{72}} = \frac {CD}{\sin{(x + 14)}}$. By dividing them we obtain: $ \sin{72}\cdot \sin{(x - 14)} = \sin{44}\cdot \sin{(x + 14)}$ $ \iff \cos{(86 - x)} = \cos{(x - 30)}\iff 2x = 116 \iff x = 58$. Thus, $ \angle{APD} = 180 - 116 = 64$., and the problem is solved.
06.05.2007 04:13
Let $A'$ be a point on $BC$ such that $AA'=AB$. Then $\angle BAA' = 180-2\cdot 44 = 92^\circ$ Similarly Let $D'$ be a point on $BC$ such that $DD'=CD$. Then $\angle CDD' = 180-2\cdot 72 = 36^\circ$ Let $E$ be the intersection point of $AA',DD'$ We have $\angle EAD = 150-92 = 58^{o}$ and $\angle EDA = 94-36 = 58^{o}$ So the triangle $EAD$ is isosceles $\Rightarrow EA=ED \ (1)$ Also $AB = CD \Rightarrow AA' = DD' \ (2)$ We'll prove that $E \in BC$. Assume that $E\not\in BC$ The line $BC$ bisects the plane into two half-planes If $E$ is on the same half-plane with $A,D$ then $EA' = AA'-EA$ and $ED' = DD'-ED$ If $E$ is on the other half-plane with $A,D$ then $EA' = EA-AA'$ and $ED' = ED-DD'$ At any case, from $(1)\wedge(2)$ we get $EA'=ED'$ which yields that $\triangle EA'D'$ is isosceles. But this is not possible because the lines $AA',DD'$ make different angles with line $BC \ \ (44^\circ \mbox{ and }72^\circ$) We got in a contradiction because we assumed that $E\not\in BC$. So the only possible case is that $E\in BC$ (this means that it is the point $P$) The segments $EA',ED'$ are equal in this case too, but this is because they have zero length So $\angle APD = 64^\circ$ Note We can generalize the problem: Theorem Let $ABCD$ be a convex quadrilateral with $A-D = 2(C-B)>0$. The perpendicular bisector of $AD$ intersects $BC$ at $P$. If $AB=CD=x$ then $x=AP=DP$ and consequently $\angle APD = 180-B-C$ Solution: Just follow the same steps (a little attention when $C>90^\circ$)