Question

Pada persegi ABCD, E dan F adalah titik-titik tengah AB dan CD. Garis AF dan CE memotong diagonal BD masing-masing di P dan Q. Buktikan 3PQ= BD.

In the square ABCD, E and F are midpoints of AB and CD respectively. Lines AF and CE meet diagonal BD in points P and Q respectively. Prove that 3PQ=BD.

Answer 1

Answer on Question #62800 – Math – Geometry

Question

Pada persegi ABCD, E dan F adalah titik-titik tengah AB dan CD. Garis AF dan CE memotong diagonal BD masing-masing di P dan Q. Buktikan 3PQ=BD.

In the square ABCD, E and F are midpoints of AB and CD respectively. Lines AF and CE meet diagonal BD in points P and Q respectively. Prove that 3PQ=BD.

Solution

If $ABCD$ is a square, then $BC = AD$, $\angle EBC = \angle FDA = 90{}^\circ$. If E and F are midpoints of AB and CD respectively, then $EB = FD$. Then $\Delta EBC \cong \Delta FDA$ by Leg-Leg (LL) Theorem.

If $ABCD$ is a square, then $AB = DC$ and $AB \parallel DC$. If E and F are midpoints of $AB$, $CD$ respectively and $AB = DC$, then $AE = CF$. If $ABCD$ is a square, then $AE \parallel CF$.

Thus, $AECF$ is a parallelogram and $EC \parallel AF$. In particular, $EQ \parallel AP$.

Consider triangle $\Delta ABP$.

By the Intercept Theorem,

$$\frac{BE}{AE} = \frac{BQ}{PQ}$$

Since $E$ is midpoint of $AB$, then $BE = AE$, and $\frac{BE}{AE} = 1$. So

$$\frac{BQ}{PQ} = 1,$$

that is,

$$BQ = PQ.$$

Consider $\Delta QCD$. By the Intercept Theorem,

$$\frac {D F}{C F} = \frac {D P}{P Q}$$

Since $F$ is midpoint of $CD$, then $CF = DF$ and $\frac{DF}{CF} = 1$. So

$$\frac {D P}{P Q} = 1$$

and $DP = PQ$.

Find $BD$:

$$BD = BQ + PQ + DP.$$

But $BQ = PQ$ and $DP = PQ$.

We finally get

$$BD = PQ + PQ + PQ = 3PQ.$$

Thus, we proved that $3PQ = BD$.

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