Question

4. Show that the ratio of the medians of two similar triangles is equal to the ratio of
their corresponding sides.​

Answer 1

$\large\underline{\bigstar \: \: {\sf SoluTion :-}}$

$\sf{Let \ \Delta ABC \cong \Delta PQR}\\\\\\\sf { \Rightarrow AD \ is \ the \ median}\\\\\\\sf {\therefore \ BD = CD = \frac{1}{2} BC}$

$\sf{Similarly, \ PS \ is \ the\ median }\\\\\\\sf {\therefore \ QS = RS = \frac{1}{2} QR}$

∠B = ∠Q {Corresponding angles of congruent triangles}

$\rm {\frac{AB}{PQ} =\frac{BC}{QR} }\\\\\\\rm {\implies \frac{AB}{PQ} =\frac{2BD}{2QS} }\\\\\\\rm {\therefore \ \frac{AB}{PQ} =\frac{BD}{QS} }$

$\sf {In \ \Delta ABD \ and \ \Delta PQS}\\\\\\\sf {\angle B = \angle Q}\\\\\\\rm {\frac{AB}{PQ}=\frac{BD}{QS}}\\\\\\\rm {\Delta ABD \cong \Delta PQS \ \{Using\ SAS\}}$

$\boxed {\sf {\therefore \ \frac{AB}{PQ} =\frac{AD}{PS}}}$

∵ ΔABC ≅ ΔPQR

If two triangles are similar, Ratio of their area = Square of the ratio of their corresponding side.

$\rm {\frac{Area \ of \ \Delta ABC}{Area \ of \ \Delta PQR} =(\frac{AB}{PQ})^{2}}\\\\\\\rm {\frac{Area \ of \ \Delta ABC}{Area \ of \ \Delta PQR} =(\frac{AD}{PS})^{2} }$

$\tt {\blacktriangleright \ \ Hence \ Proved }$