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$(1)$$AB$ and $CD$ are two intersecting line segments, and $P$,$Q$ are their respective midpoints. If $AB$ bisects $\angle CPD$ and $PA^2=PB^2=PC.PD$, then prove that the points $A,B,C$ and $D$ are concyclic.

I have a proof of the converse of the above theorem. Consider four concyclic points $A(a),B(b),C(c)$ and $D(d)$. Let $O$ be the midpoint of $CD$ . Then the cross ratio of these points is, $$ \lambda=\frac {AC.BD}{AD.BC} $$ $$ = \frac {(a-c)(b-d)}{(a-d)(b-c)}=-1 $$ Rearranging we get: $$ (a+b)(c+d)=2(ab+cd) $$ Rearranging again, we get: $$ \left\{a-\frac {1}{2}(c+d)\right\}\left\{b-\frac {1}{2}(c+d)\right\}=\left\{\frac {1}{2}(c-d)\right\}^2 $$ This implies: $$ OA.OB=OC^2=OD^2 $$ That is,$OA$and $OB$ are equally inclined to $CD$. $$ { } $$ I could have written all of these lines in reverse, but I don't think that would be just, for both the theorems are converses of each other. Is there a distinct proof of theorem $(1)$ either using Euclidian geometry or complex numbers (other than this way)? Any help would be appreciated.

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Let $CD\cap CD=\{E\}$, $E$ placed between $P$ and $B$ and between $C$ and $Q$.

Thus, since $PE$ is a bisector of $\Delta PCD$, we obtain: $$PE^2=PC\cdot PD-CE\cdot ED=PA^2-CE\cdot ED.$$ Id est, $$CE\cdot ED=PA^2-PE^2=(PA-PE)(PA+PE)=BE\cdot AE,$$ which gives $$\frac{CE}{AE}=\frac{EB}{ED},$$ which says $$\Delta CEB\sim\Delta AED,$$ which gives $$\measuredangle BCD=\measuredangle BAD$$ and quadrilateral $ACBD$ is cyclic.

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$$\frac{PC}{PB}=\frac{PB}{PD}, \ \angle CPB=\angle BPD \implies \triangle CPB\sim \triangle BPD$$

These triangles have the same angles ($\alpha,\beta,\gamma$) as shown.

$$\frac{PC}{PA}=\frac{PA}{PD}, \ \angle CPA=\angle APD \implies \triangle CPA\sim \triangle APD$$

These triangles have the same angles ($\delta,\varphi,\varepsilon$) as shown.

In quadrilateral $ABCD:$

$$\angle C= \varphi+\gamma$$

$$\angle D= \beta+\varepsilon$$

$$\angle C +\angle D=\varphi+\gamma + \beta+\varepsilon=180^\circ$$

The last is obvious from triangle $ABC$. Sum of opposite angles is $180^\circ$ and therefore the quadrilateral is concyclic.

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