Question
In Fig. $\text{AD}\perp\text{CD}$ and $\text{CB}\perp\text{CD}.$ If $\text{AQ}=\text{BP}$ and $\text{DP}=\text{CQ},$ prove that $\angle\text{DAQ}=\angle\text{CBP}.$
✓
Answer
In $\triangle\text{DAQ}$ and $\triangle\text{CBP}$$\angle\text{ADQ}=\angle\text{BCP}=90^\circ$
$\text{DP}=\text{CQ}$ [given]
$\Rightarrow\text{DP}+\text{PQ}=\text{CQ}+\text{PQ}$
$\Rightarrow\text{DQ}=\text{CP}$
$\text{AQ}=\text{BP}$ [given]
By RHS congurence criterion $\triangle\text{DAQ}\cong\triangle\text{CBP}$$\therefore\angle\text{DAQ}=\angle\text{CBP}$ [c.p.c.t]
$\text{DP}=\text{CQ}$ [given]
$\Rightarrow\text{DP}+\text{PQ}=\text{CQ}+\text{PQ}$
$\Rightarrow\text{DQ}=\text{CP}$
$\text{AQ}=\text{BP}$ [given]
By RHS congurence criterion $\triangle\text{DAQ}\cong\triangle\text{CBP}$$\therefore\angle\text{DAQ}=\angle\text{CBP}$ [c.p.c.t]
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