## BdMO National Higher Secondary 2019/9

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### BdMO National Higher Secondary 2019/9

Let \$ABCD\$ is a convex quadrilateral.The internal angle bisectors of \$\angle {BAC}\$ and \$\angle {BDC}\$ meets at \$P\$.\$\angle {APB}=\angle {CPD}\$.Prove that \$AB+BD=AC+CD\$.

Abdullah Al Tanzim
Posts: 24
Joined: Tue Apr 11, 2017 12:03 am

### Re: BdMO National Higher Secondary 2019/9

Let \$B'\$ be a point on the line \$AB\$ such that \$AB'=AC\$ and \$C'\$ be a point on the line \$DC\$ such that \$DC'=BD\$.
So, it suffices to prove that \$BB'=CC'\$.

\$\triangle AB'P \cong \triangle ACP\$ \$ \Rightarrow \angle APB'=\angle APC\$ \$ \Rightarrow \angle BPB'=\angle APD\$.

Similiarly, \$\triangle DBP \cong \triangle DC'P \Rightarrow \angle CPC'=\angle APD \$
So, \$\angle BPB'=\angle CPC'\$.

In \$ \triangle BPB'\$ and \$ \triangle CPC',\$
\$\angle BPB'=\angle CPC'\$
\$BP=PC'\$
\$B'P=PC\$
So, \$\triangle BPB' \cong \triangle CPC'\$
So, \$BB'=CC' \Rightarrow AB+BD=AC+CD\$
\$(Q.E.D.)\$
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO National Higher Secondary 2019/9

No need to draw a figure?