Please help with this proof! Itwould really help if oyu drew a diagram, thanks!
ABC has a right angle at B. Legs AB and CB are extended past point B to points D and E, respectively, such that <EAC=<ACD=90*. Prove that EB * BD = AB * BC
+130511 Here's a pic of the situation :
m< EAB + m< BAC = 90 given
Since ABC is a right angle, then so is ABE
Then m<EAB + m<BEA = 90 [ three angles of any triangle sum to 180 ]
Then m< EAB + m< BAC = m<EAB + m<BEA [ subtract m<EAB from both sides]
m< BAC = m <BEA
And this means that m<ACB = m<BAE
Then, by AAA, ΔEBA is similar to ΔABC
And in a like manner....we can prove that ΔCBD is similar to ΔABC
But ΔEBA was already proved to similar ΔABC as well ....so....
ΔCBD is also similar to ΔEBA
Thus AB = DB
BE BC
And, by cross-product equality..... BE * DB = AB * BC
And BE = EB and DB = BD
So...... EB * BD = AB * BC
