In the given figure ABCD is a quadrilateral in which AB is parallel. DC andP is the midpoint of BC .on producing AP and DC meet at Q .prove that AB=CQ. DQ=DC +AB
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Given: ABCD is a Quadrilateral
AB parallel DC
BP =PC
To prove: AB=CQ
DQ= DC+AB
Proof: In Triangle ABP and Triangle ACQ ,We have
AP= PQ (COMMON)
AB Parallel AC is a transversal
Angle(BAP)=Angle(PQC) (ALT.INT. ANGLE)
Angle(BPA)=Angle(CPQ) (VER.OPP. ANGLE)
Triangle APB~ Triangle QPC (A.S.A)
AB=CQ (C.P.C.T)
DQ = DC + CQ
DQ = DC + AB
![hence \: proof hence \: proof](https://tex.z-dn.net/?f=hence+%5C%3A+proof)
AB parallel DC
BP =PC
To prove: AB=CQ
DQ= DC+AB
Proof: In Triangle ABP and Triangle ACQ ,We have
AP= PQ (COMMON)
AB Parallel AC is a transversal
Angle(BAP)=Angle(PQC) (ALT.INT. ANGLE)
Angle(BPA)=Angle(CPQ) (VER.OPP. ANGLE)
Triangle APB~ Triangle QPC (A.S.A)
AB=CQ (C.P.C.T)
DQ = DC + CQ
DQ = DC + AB
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