Question

prove that corresponding medians of two similar triangles are proportional to the corresponding sides of the triangle
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The proof is explained below :

Step-by-step explanation:

$\text{To Show}: \frac{AD}{PM}= \frac{AB}{PQ}= \frac{BC}{QR}= \frac{AC}{PR}\\\\\text{Step 1 : }\Delta ABC\sim\Delta PQR\\\text{And we know that the sides of similar triangles are proportional to each other}\\\text{and corresponding angles are equal}\\\\\implies\frac{AB}{PQ}= \frac{BC}{QR}= \frac{AC}{PR}\thinspace and\thinspace\angle B\cong\angle Q$

$\text{Step 2 : }\frac{AB}{PQ}= \frac{BC}{QR}= \frac{2\cdot BD}{2\cdot QM}\text{ ; P and M are respective mid points}\\\\\implies \frac{AB}{PQ}= \frac{BD}{QM}$

$\text{Step 3 : In }\Delta ABD\thinspace and\thinspace\Delta PQM,\\\\\frac{AB}{PQ}= \frac{BD}{QM}\\\\\angle B\cong\angle Q\text{ ; By step 1}\\\text{Now, By SAS postulate of similarity of triangles }\\\Delta ABD\sim \Delta PQM\\\implies \frac{AB}{PQ}= \frac{BD}{QM}= \frac{AD}{PM}$

$\text{Step 4 : So by all the above steps we conclude}\\\frac{AD}{PM}=\frac{AB}{PQ}= \frac{BC}{QR}=\frac{AC}{PR}$

So, we can say that corresponding medians of two similar triangles are proportional to the corresponding sides of the triangle.

Hence Proved.