問題:
題解:
Joint BT and DT.
Let MT=a, ME=b, MD=r and MB=s
Our aim is to prove a=b.
Consider △BMT and △TMD,
∠BMT=∠TMB (commond ∠)
∠TBM=∠DTM (∠ in alt. segment)
Thus, △BMT∼△TMD (AA)
By corr. sides ∼△s, we have,
MTMD=MBMTar=saa=√rs
Consider △BME and △EMD,
∠BME=∠EMD (commond ∠)
∠MBE=∠A (∠s in the same segment)
∠A=∠MED (alt. ∠s, EM//AC)
so, ∠MBE=∠MED
Thus, △BME∼△EMD (AA)
By corr. sides ∼△s, we have,
MEMD=MBMEbr=sbb=√rs=a
Therefore, MT=ME.
Q.E.D.
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