Question

in △lmn, the angle bisector of ∠l intersects mn at p. if lp is the angle bisector and mp = 9, pn = 6, and lm = 18, what is the length of ln?
options:
a. 13
b. 12
c. 15
d. 14

if a triangle is divided into two smaller triangles by an altitude and all three triangles are similar, what theorem could this be used to prove?
options:
a. sas criterion
b. pythagorean theorem
c. sss criterion
d. hl criterion

in △abc, ∠acb = 90°, and cd is the altitude to ab. which of the following is true?
options:
a. cd is proportional to the sum of ac and bc
b. the altitude divides ab into two congruent segments
c. △acd and △bcd are congruent to each other
d. △acd ~ △abc ~ △bcd

Explanation:

First Question:

Step1: Apply Angle Bisector Theorem

The Angle Bisector Theorem states that if a bisector of an angle of a triangle divides the opposite side into segments proportional to the adjacent sides. So, $\frac{LM}{LN} = \frac{MP}{PN}$.

Step2: Substitute known values

We know $LM = 18$, $MP = 9$, and $PN = 6$. Substitute these into the proportion: $\frac{18}{LN} = \frac{9}{6}$.

Step3: Solve for LN

Cross - multiply: $9\times LN=18\times6$. Then $9LN = 108$. Divide both sides by 9: $LN=\frac{108}{9}=12$.

When a right triangle is divided into two smaller right triangles by an altitude to the hypotenuse, all three triangles (the original and the two smaller ones) are similar. This property is used to prove the Pythagorean Theorem. The SAS, SSS, and HL criteria are for triangle congruence or similarity in different contexts, not related to this proof of the Pythagorean Theorem.

In right triangle $ABC$ with right angle at $C$ and altitude $CD$ to hypotenuse $AB$:

• Option a: The length of the altitude $CD=\frac{AC\times BC}{AB}$, it is not proportional to the sum of $AC$ and $BC$.
• Option b: The altitude divides $AB$ into two segments $AD$ and $DB$ such that $AD=\frac{AC^{2}}{AB}$ and $DB = \frac{BC^{2}}{AB}$, they are congruent only if $AC = BC$, which is not generally true.
• Option c: $\triangle ACD$ and $\triangle BCD$ are similar, not congruent (unless $AC = BC$).
• Option d: By the property of right triangles and altitudes, $\triangle ACD\sim\triangle ABC$ (AA similarity, $\angle A$ is common and both are right triangles) and $\triangle BCD\sim\triangle ABC$ (AA similarity, $\angle B$ is common and both are right triangles), so $\triangle ACD\sim\triangle ABC\sim\triangle BCD$.

Answer:

b. 12

Second Question: