It is known that, as shown in the figure, △ ABC is an equilateral triangle, AE = CD, BQ ⊥ ad at Q, be crossing ad at point P, Confirmation: BP = 2pq

It is known that, as shown in the figure, △ ABC is an equilateral triangle, AE = CD, BQ ⊥ ad at Q, be crossing ad at point P, Confirmation: BP = 2pq

It is proved that ∵ △ ABC is an equilateral triangle,
∴AB=AC，∠BAE=∠C=60°，
In △ ABE and △ CAD,
AB＝AC
∠BAE＝∠C＝60°
AE＝CD ，
∴△ABE≌△CAD（SAS），
∴∠1=∠2，
∴∠BPQ=∠2+∠3=∠1+∠3=∠BAC=60°，
∵BQ⊥AD，
∴∠PBQ=90°-∠BPQ=90°-60°=30°，
∴BP=2PQ．

As shown in the figure, it is known that in the triangle ABC, D and E are points on edge AB and AC respectively, and De is parallel to BC, ad is the proportion median of AF and AB, and point F is on edge ab, Verification: fed = ∠ DCB

prove:
∵DE∥BC
∴AD/AB=AE/AC,∠DCB=∠CDE
∵ ad is the median ratio of AB and AF
∴AF/AD=AD/AB
∴AF/AD=AE/AC
∴EF∥CD
∴∠FED=∠CDE
∴∠FED=∠DCB
The math group answered your question,

As shown in the figure, in the triangle ABC, the angle BAC = 90 degrees, ab = AC, ad vertical, BC and D, e are points on AC, be intersects ad with H, AF is perpendicular to be to g, Confirmation: DH = DF With the knowledge of grade one

Because the angle BAC = 90 degrees, ab = AC, ad vertical BC
So BD = ad (Pythagorean theorem and isosceles triangle in one)
Because ad vertical BC, AF vertical be
So angle BDA = angle ADC (vertical definition)
Angle BFA = angle Bhd
So triangle BDH is equal to triangle ADF (A.A.S.)
So DH = DF
Congruent triangle, as I remember, was taught in the first grade of junior high school

As shown in the figure, △ ABC is an isosceles right triangle, ∠ ACB = 90 ° and ad is the center line on the side of BC. Through C, the vertical line of ad is made, which intersects AB at point E and ad at point F. it is proved that ∠ ADC = ∠ BDE

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∵AD⊥BC
∴∠BAD+∠B=90°
∵∠1=∠B
∴∠1+∠BAD=∠BAC=90°
The △ ABC is a right triangle

As shown in the figure, the area of the triangle ABC is 54 square centimeters, be: EC = 1:2, ad: DB = 1:2. Find the area of the triangle ade. (AE is the BC vertical line.) The answer should be 12

The answer is 6. By drawing and calculating by myself, we know that the area of ade of triangle is one third of Abe, Abe is one third of big triangle, that is, ade is one ninth of big triangle

As shown in the figure, in △ ABC, ab = 12cm, AE = 6cm, EC = 4cm, and AD BD＝AE EC． ① To find the length of AD AB＝EC AC．

① Let ad = xcm, then BD = ab-ad = (12-x) cm
∵AD
BD＝AE
EC，
∴x
12−x＝6
Four
The solution is x = 7.2cm
∴AD=7.2cm；
②∵AD
BD＝AE
EC，
∴AD+BD
BD＝AE+EC
EC
That is ab
BD＝AC
EC．
∴BD
AB＝EC
AC．

In isosceles △ ABC, ab = AC, ad ⊥ BC on D, point E on line AC, CE = 1 2Ac, ad = 18, be = 15, then the area of △ ABC is______ .

As shown in the figure, ∵ in the isosceles ⊥ ABC, ab = AC, ad ⊥ BC in D, ∵ ad is the center line of the bottom BC,

As shown in the figure, point E is outside △ ABC, point D is on the edge of BC, de intersects AC at point F, if ∠ 1 = ∠ 2 = ∠ 3, AC = AE, verification: ab = ad

It is proved that: ∵ 1 = ∠ 2,
∴∠1+∠DAF=∠2+∠DAF，
That is ∠ BAC = ∠ DAE,
∵∠2=∠3，∠AFE=∠DFC，
∴∠E=∠C，
In △ ABC and △ ade,
∠BAC＝∠DAE
∠E＝∠C
AC＝AE ，
∴△ABC≌△ADE（AAS），
∴AB=AD．

As shown in the figure, be ⊥ ad, CF ⊥ AD are known, and be = cf. please judge whether AD is the center line of ⊥ ABC or the angular bisector? Please state the reasons for your judgment

Ad is the midline of △ ABC
The reasons are as follows.
∵BE⊥AD，CF⊥AD，
∴∠BED=∠CFD=90°，
In △ BDE and △ CDF,
∠BED＝ ∠CFD
∠BDE＝∠CDF
BE＝CF
∴△BDE≌△CDF（AAS），
∴BD=CD．
The ad is the midline of △ ABC