In right trapezoid ${ABCD \left(AB\parallel CD\right)}$ the angle at vertex $B$ measures ${{75}^{{}^\circ }}$. Point ${H}$is the foot of the perpendicular from point ${A}$ to the line ${BC}$. If ${BH=DC}$ and${AD+AH=8}$, find the area of ${ABCD}$.
Problem
Source: 2009 JBMO Shortlist G2
Tags: geometry, JBMO
ChopanovDovlet
02.02.2019 17:57
Give me solution
Steve12345
02.02.2019 18:01
@above https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2016/07/shl-2009.pdf
ChopanovDovlet
02.02.2019 18:04
Please give me solution I solved this problem but It was wrong =9,2...
Steve12345
02.02.2019 18:10
The answer is 8. If you press the link you will see the official solution(Or just google JBMO 2009 shortlist )
Lemmas
05.05.2023 19:57
ChopanovDovlet wrote: Give me solution Let P be a intersection of BC and AD. Obviously tr. DPC=tr. HAB --> S(ABCD)=S(APH) . AD+AH=AP=8 and angle APH=15°. We have obvious fact. If HH' is height in tr. APH then HH'=AP/4=2. So S(ABCD)=S(APH)=8