Question

36. In the figure triangleABC~ triangle DEF and CM and FN are respectively the bisectors of angle ACB
and angle DFE. Prove that
i) triangle AMC~triangle DNF
ii)triangle BCM~triangle EFN
ii) CM/FN = AC/ DF​

Answer 1

Hi,

Answer:

Since ∆ABC & ∆DEF are similar to each other, then by SSS and SAS we can say

$\frac{AB}{DE}$ = $\frac{BC}{EF}$ = $\frac{AC}{DF}$ …. (i)

And,  

∠A = ∠D …. (ii)

∠B = ∠E

∠C = ∠F

We are also given that CM and FN are angle bisectors from ∠ACB and ∠DEF respectively. So, by angle bisector theorem we know that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle i.e., we have  

In ∆ABC,

$\frac{AC}{BC}$ = $\frac{AM}{BM}$ ….. (iii)

And,  

In ∆DEF,

$\frac{DF}{EF}$ = $\frac{DN}{EN}$ …. (iv)

From (i), (iii) & (iv), we can also write  

$\frac{AC}{BC}$ = $\frac{AM}{BM}$ = $\frac{DF}{EF}$ = $\frac{DN}{EN}$  

Or, $\frac{AM}{DN}$ = $\frac{AC}{DF}$ ….. (v)

And, $\frac{EN}{BM}$ = $\frac{EF}{BC}$ ….. (vi)

1. Since we have  

$\frac{AM}{DN}$ = $\frac{AC}{DF}$ [from (v)]

∠A = ∠D [from (ii)]

By SAS

ΔAMC ~ ΔDNF, hence proved

2.Similarly, we also have  

$\frac{EN}{BM}$ = $\frac{EF}{BC}$ and ∠B = ∠E

By SAS

ΔBCM ~ ΔEFN, hence proved

3. From above we got ΔAMC ~ ΔDNF, so by SSS we can say

$\frac{AM}{DN}$ = $\frac{AC}{DF}$ = $\frac{CM}{FN}$

Or, $\frac{CM}{FN}$ = $\frac{AC}{DF}$, hence proved  

Hope this helps!!!!