Given: LH ⊥ EH , PK ⊥ EH
Prove: EK/EH=EP?EL
To prove that EK/EH = EP/EL, we can use the properties of similar triangles.
First, let's consider triangle EKH and triangle EPH. These two triangles share an angle, E, and have a right angle at H. We can deduce that they are similar by the AAA (Angle-Angle-Angle) similarity criterion.
Since they are similar triangles, the corresponding sides are proportional. Therefore, we can write the proportion:
EK/EH = EP/EP
Simplifying the proportion, we have:
EK/EH = 1
Now, let's consider triangle ELH and triangle EPH. Just like before, these two triangles share an angle, E, and have a right angle at H. By the AAA similarity criterion, we can conclude that they are also similar triangles.
Using the same logic, we can write the proportion:
EL/EH = EP/EP
Simplifying the proportion, we have:
EL/EH = 1
Now, let's compare our two proportions:
EK/EH = EP/EP
EL/EH = 1
From the second proportion, we can rearrange it to get EL = EH.
Now, we can substitute this value in the first proportion:
EK/EH = EP/EP
Since EL = EH, we can substitute EL for EH:
EK/EL = EP/EP
Because any number divided by itself is always 1, we can simplify the equation to:
EK/EL = 1
Now, observe that if EK/EL = 1, we can conclude that EK = EL.
So, we have:
EK = EL
Now, we can use this equality in the proportion we obtained earlier:
EK/EH = EP/EP
Since EK = EL, we have:
EL/EH = EP/EP
Finally, recall that we previously established that EL = EH. Therefore:
EH/EH = EP/EP
Again, any number divided by itself is always 1, so we simplify the equation to:
1 = 1
Thus, we have proved that EK/EH = EP/EL.
you really think that will get you a answer? Give some context.
typos and no context.
Take a hike.