Deromediqm

Answered

2022-07-23

In $\mathrm{\u25b3}ABC$, let D be a point on BC such that AD bisects $\mathrm{\angle}A$. If $AD=6$, $BD=4$ and $DC=3$, then find AB.

Answer & Explanation

edgarovhg

Expert

2022-07-24Added 12 answers

Step 1

Let E is point of line AD such that $\mathrm{\angle}AEB=90\xb0$ and F is point of line AD such that $\mathrm{\angle}AFC=90\xb0$. Then triangles ABE and ACF are similar, triangles DBE and DCF are similar. Using this facts, one can write $\frac{AF}{AE}=\frac{CF}{BE}=\frac{CD}{BD}=\frac{3}{4}$

$\frac{DF}{DE}=\frac{CD}{BD}=\frac{3}{4}$

Step 2

Then $\frac{EF}{AE}=1-\frac{AF}{AE}=\frac{1}{4}$

$\frac{EF}{DE}=1+\frac{DF}{DE}=\frac{7}{4}$

$\frac{AE}{DE}=\frac{EF/DE}{EF/AE}=7$

$\frac{AD}{DE}=\frac{AE}{DE}-1=6$

$DE=AD/6=1$

$AE=7\cdot DE=7$

$B{E}^{2}=B{D}^{2}-D{E}^{2}=16-1=15$

$A{B}^{2}=A{E}^{2}+B{E}^{2}=49+15=64$

$AB=8$

Let E is point of line AD such that $\mathrm{\angle}AEB=90\xb0$ and F is point of line AD such that $\mathrm{\angle}AFC=90\xb0$. Then triangles ABE and ACF are similar, triangles DBE and DCF are similar. Using this facts, one can write $\frac{AF}{AE}=\frac{CF}{BE}=\frac{CD}{BD}=\frac{3}{4}$

$\frac{DF}{DE}=\frac{CD}{BD}=\frac{3}{4}$

Step 2

Then $\frac{EF}{AE}=1-\frac{AF}{AE}=\frac{1}{4}$

$\frac{EF}{DE}=1+\frac{DF}{DE}=\frac{7}{4}$

$\frac{AE}{DE}=\frac{EF/DE}{EF/AE}=7$

$\frac{AD}{DE}=\frac{AE}{DE}-1=6$

$DE=AD/6=1$

$AE=7\cdot DE=7$

$B{E}^{2}=B{D}^{2}-D{E}^{2}=16-1=15$

$A{B}^{2}=A{E}^{2}+B{E}^{2}=49+15=64$

$AB=8$

equissupnica7

Expert

2022-07-25Added 4 answers

Step 1

With angle bisector theorem you can easily find out that $\frac{AB}{AC}=\frac{4}{3}$. By this relationship, you can label sides $AB=4x,AC=3x$.

Draw perpendicular DK, DL such that points D, L lies on the sides AB, AC respectively.

You can easily find out that $\mathrm{\u25b3}AKD\cong \mathrm{\u25b3}DLA$ (A. A. S.)

By this, you can label $KD=DL=h$ and $AK=AL=a$

Step 2

By the Pythagorean theorem, you can get the following equations,

${h}^{2}+{a}^{2}=36$ (1)

${h}^{2}+(3x-a{)}^{2}=9$ (2)

${h}^{2}+(4x-a{)}^{2}=16$ (3)

Expanding (2), (3) and substituting (1) as required will lead you to $x=2$

As $AB=4x,AB=2\ast 4=8$

With angle bisector theorem you can easily find out that $\frac{AB}{AC}=\frac{4}{3}$. By this relationship, you can label sides $AB=4x,AC=3x$.

Draw perpendicular DK, DL such that points D, L lies on the sides AB, AC respectively.

You can easily find out that $\mathrm{\u25b3}AKD\cong \mathrm{\u25b3}DLA$ (A. A. S.)

By this, you can label $KD=DL=h$ and $AK=AL=a$

Step 2

By the Pythagorean theorem, you can get the following equations,

${h}^{2}+{a}^{2}=36$ (1)

${h}^{2}+(3x-a{)}^{2}=9$ (2)

${h}^{2}+(4x-a{)}^{2}=16$ (3)

Expanding (2), (3) and substituting (1) as required will lead you to $x=2$

As $AB=4x,AB=2\ast 4=8$

Most Popular Questions